### I'm not a PhD but...

by why_bird (Pilgrim)
 on Feb 03, 2009 at 08:19 UTC ( #740912=note: print w/replies, xml ) Need Help??

in reply to Re^3: How many colors does a rainbow have?
in thread How many colors does a rainbow have?

But there are also colors in the infrared and ultraviolet present as well.

Isn't this a contradiction? Colo[u]r being those frequencies of light visible to humans... and infra-red and ultra-violet not being among those...?

While each atom in the photosphere may emit light at one quantum frequency, the sun is so hot that doppler shifting of the light causes the lines to "fuzz out" so that you see essentially a continuous spectrum of light.

I might be wrong about this, but isn't the vast majority of the sun's radiation thermal, i.e. blackbody, i.e. continuous in the frequency domain? Yes, there are spectral lines caused by atomic de-excitation, and they are broadened by Doppler shifting due to the temperature of the gas in the sun (amongst other things) but this isn't what the majority of radiation emitted from the sun is.. is it?

Someone help me out here!
why_bird
........
Those are my principles. If you don't like them I have others.
-- Groucho Marx
.......

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Re: I'm not a PhD but...
by jdporter (Canon) on Feb 03, 2009 at 15:11 UTC
Isn't this a contradiction? Colo[u]r being those frequencies of light visible to humans...

All of these arguments depend on the definition of "color", and until we all agree on one, there will never be agreement on how many "colors" are in a rainbow.

isn't the vast majority of the sun's radiation thermal, i.e. blackbody

Well, obviously, essentially all of the sun's radiation is blackbody, in that none of it is reflected or transmitted. :-)

blackbody, i.e. continuous in the frequency domain

While blackbody radiation may be continuous theoretically, that doesn't mean that thermal radiators like the sun have continuous spectra in reality.

Brass tacks: Is the solar spectrum continuous or quantified? If the physics PhD says it's continuous, I'll believe him.

Yes, there are spectral lines ... but this isn't what the majority of radiation emitted from the sun is.. is it?

Is there any reason to suppose that the "smoothness" of the solar spectrum is qualitatively different from one part to another? No; In fact, it looks like the roughness is merely proportional to the intensity, across the spectrum. It also appears that whatever "fuzzing out" is happening, it is not enough to completely smooth out the spectrum. Far from it!

Between the mind which plans and the hands which build, there must be a mediator... and this mediator must be the heart.

Thanks for answering my post jdporter. You've let yourself in for it a bit though, as you've generated more questions :) If you or anyone else can help me understand this I'd be grateful. I realise it's hideously OT in terms of perl, so feel free to ignore me, but here goes(!)

Whilst I agree that without a definitive definition of 'colour' (or 'color' :P) we won't agree how many colours are in the rainbow, I am proposing that part of the definition of 'colour' should include 'wavelengths of light that humans can see'. By definition (at least any that I've ever seen) infra-red and ultra violet are outisde the visible spectrum. That's all :)

As to the rest of my post.. I am bad at expressing myself with physics! Let me try again.. for one it should have sounded more like a (series of?) question(s):

Firstly, am I right in thinking that a blackbody doesn't just require not to 'reflect or transmit' (I am not quite sure what you mean by this) any of it's radiation? It must also have a continuous spectrum which is dependent only on temperatue, and is determined by Planck's law? And it must be able to absorb and emit radiation at any (and all) wavelengths?

Given that the Sun is (approximately at least) a blackbody radiator, doesn't that mean that it's spectrum is continuous, and it (absorbs and) radiates EM waves at all frequencies? If the previous is correct, is the mechanism for the sun's radiation atomic exctitation/de-excitation? Or plasma recombination, I would've thought more likely.

If the main mode of EM emission by the sun is through atomic excitation/de-excitation, is that true of all black bodies? What other modes (if any) of photon emission are there from materials? (Thinking about it, the only other ones I know would be the acceleration of charged particles---relevant to a plasma? and annihiliation of matter-antimatter pairs---plausible in something pretty hot I would have thought). If so, how is it that a theoretical black body has a continuous spectrum, dependent only on temperature if the main mode of anything's EM emission is through de-excitation? Wouldn't you expect the spectrum to be concentrated around spectral lines?

It also appears that whatever "fuzzing out" is happening, it is not enough to completely smooth out the spectrum. Far from it!

My point is though, shouldn't it? If the main mode of EM emission from a black body is by atomic excitation/de-excitation, shouldn't black body theory take account of this, or use it as its starting point?

Thanks!
why_bird
........
Those are my principles. If you don't like them I have others.
-- Groucho Marx
.......
It must also have a continuous spectrum which is dependent only on temperatue, and is determined by Planck's law?

It must, but the "continuous" part is mis-leading.

Planck's perfect black body is - like everything in physics - just a model, and contains assumptions and approximations.

The assumptions are that if your body has a length L, then the wavelengths lambda all have the property L = n * lambda, where n is an integer, (and that all absorption and transmission happens between two sharp energy levels; not 100% about this one).

The simplification is that in order to calculate the spectral power distribution, one replaces a sum by an integral, and thus looses the quantization condition on the way.

So in the model of the Planck black body radiator the continuous spectrum is just an artifact of a mathematical approximation, not a physical property.

Note that there are other mechanism that take care of smoothing the spectrum. For example all particles in the sun move (brownian motion), so the radiation has a red- or blue shift. The second mechanism is that all processes - including those of photon emissions - take finite time, so that have (by virtue of the Heisenberg uncertainty principle) a non-sharp energy distribution (called "intrinsic" line width)

My point is though, shouldn't it? If the main mode of EM emission from a black body is by atomic excitation/de-excitation, shouldn't black body theory take account of this, or use it as its starting point?

I'm not too firm with the astronomy of our sun, so I could be wrong, but... I think the "black body" approximation holds true for the bulk of the sun, but not for the out regions (Corona, whatever), and these outer regions could be enough to filter all the spectral lines that you see when looking at the sun.

You're trying to mix two different models (atomic spectra vs. black body, a thermodynamic model), so you have to be very careful which part of which model still holds true in the mixed case.

Re: I'm not a PhD but...
by jonadab (Parson) on Feb 04, 2009 at 12:40 UTC
Colo[u]r being those frequencies of light visible to humans...

If you define colour in terms of what the human eye sees, there are only three of them, four if you count luminosity that's not in any of the three color categories, or throw in tetrochromacy; five if you count both. So yellow, for instance, is not a color if you define it this way. It's a combination of colours (specifically, red and green), or else it's a pigment that absorbs a certain color (blue).

And yeah, the output of the sun is pretty well continuous across the electromagnetic spectrum, so if you define colors as wavelengths of electromagnetic radiation then the cardinality of the set of all colours in the rainbow is aleph-sub-one, the same as the cardinality of the set of all real numbers. I suppose that means the number of colors in the rainbow is actually greater than infinity, if you define infinity in the usual gradeschool way (which comes out in math as aleph-sub-naught, the cardinality of the set of natural counting numbers).

If you define colour in terms of what the human eye sees, there are only three of them

I think you're misconstruing me! I am limiting colours to be 'things which humans can see' which doesn't correspond to 'the specific frequency ranges which stimulate only one cone type'. (And in fact the frequency ranges overlap so your definition becomes a bit more tricky than you make out)

I can see EM radiation with wavelength 570nm (yellow), so I would define it as a colour! I just happen to see it by stimulating more than one cone type at once.

so if you define colors as wavelengths of electromagnetic radiation then the cardinality of the set of all colours in the rainbow is aleph-sub-one

As tilly pointed out, you don't get mixtures of blue and red light in a rainbow, which we see as pink-purples (the line of purples on the CIE chromacity diagram.)

Update: In fact, you don't get any mixing in a rainbow (approximately, obv mist is not a perfect refractor and probably some other caveats like angle and distance), so it's like going around the edge of the chromacity diagram... so there's quite a few less than all possible colours/hues in a rainbow.. but still probably infinite (I'm not a mathematician either, so no fancy 'alehps' from me... :P). That's my current conjecture anyway!
........
Those are my principles. If you don't like them I have others.
-- Groucho Marx
.......
As tilly pointed out, you don't get mixtures of blue and red light in a rainbow, which we see as pink-purples (the line of purples on the CIE chromacity diagram.)

It's way more complicated than that. What we see at any given point, at any given moment in time, in our mental image of a rainbow, is not the result of a single pure frequency stream of EM emanating from that point.

If that were the case, then everything we see would scintillate between shades of red, blue and green as we move our heads. Because if the stream of photons coming from that point source are accurately focused, they would only hit one cone at a time: say red. Then when you moved your head slightly, that point source would hit a different cone, say green. And so on.

You have to appreciate that the image we see is a complex, time-averaged (think:persistance of vision), sum of the frequency responses of photons hitting many cones of the three types, as that point source and/or our heads and bodies move. In the same way that ultrasound images are not instantaneous snapshots of the reflected waves from a single position of the tranducer, but rather a complex mapping of the responses from many points as the transducer is moved around in both 3D space, and over time.

So whilst there my not be any mixtures of 'red' and 'blue' frequencies in the spectrum at the source of the light, by the time it has been refracted through a billion drops of water, with some frequencies being attuenuated by the gasses in the air on the way to and from the raindrops; and by the water, and whatever contaminants it contains, of the raindrop itself. Add to that, any coincident (reflected, refracted and backgound light from other sources), photons that reach our eyes from the same direction, and what we "see" is an entirely different thing to what is there (in the rainbow), to begin with.

Even teh rainbow itself is an entirely nebulous entity. Consisting of only some small patr of the frequencies in the original light source that happen to refract in out direction. If air currents (wind, inversion layers etc.), cause the falling raindrops to change tragectory half way through their transition across the piece of sky where the angle is correct for the light to be refracted in our direction, then the frequency of the portion of the source that comes our way will change. Simple put, the rainbow will appear to ripple.

The term "colo(u)r" only makes sense in terms of our perception of what we see. And that perception is far from set in the frequencies of EM coming from any given point source. It is also influenced (a lot) by everything else we are seeing at the time. That is no more clearly demonstrated than by this famous optical illusion

Our perceptions have almost nothing to do with the frequencies of the EM spectrum. Hence my brother's yellowish coloured car looked almost purple under low-pressure sodium street lighting.

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

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